--- title: "Orthogonality of modes and modal analysis" subject: "Theory of Mechanism and Machine II" chapter: 8 key: "semester_7/theory_of_mechanism_and_machine_ii:orthogonality-modal-analysis-forced-response" source: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/orthogonality-modal-analysis-forced-response site: BCT Engineering --- # Orthogonality of modes and modal analysis Theory of Mechanism and Machine II · Chapter 8: Vibration of Multi Degree of Freedom Systems - Asked in: year not given, 6 marks - Video: [Part-3 । Principle of orthogonality of Modes । Modal Analysis । MDFS Vibration Engineering ।](https://www.youtube.com/watch?v=vFI0a3mYhX0) — Mechanical Engineering Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/theory_of_mechanism_and_machine_ii/orthogonality-modal-analysis-forced-response - Whole chapter: https://www.bctengineer.com/md/semester_7/theory_of_mechanism_and_machine_ii/chapter-8 ## Question Prove that the natural modes of an undamped multi-degree-of-freedom system are orthogonal with respect to the mass and stiffness matrices. Explain how modal analysis uses this property to find the response to a general forcing function. ## Short answer For distinct natural frequencies, symmetry of mass and stiffness matrices leads to {φ}_r^T[M]{φ}_s=0 and {φ}_r^T[K]{φ}_s=0. Modal analysis uses the mode shape matrix to uncouple the equations via orthogonality, reducing the system to independent SDOF equations; each is solved by convolution, then transformed back to physical coordinates. ## Answer #### Orthogonality Let $\omega_r, \{\phi\}_r$ and $\omega_s, \{\phi\}_s$ be two natural frequencies with their mode shapes, with $\omega_r \ne \omega_s$. Then $$ [K]\{\phi\}_r = \omega_r^2[M]\{\phi\}_r \quad (1), \qquad [K]\{\phi\}_s = \omega_s^2[M]\{\phi\}_s \quad (2) $$ Pre-multiply (1) by $\{\phi\}_s^T$ and (2) by $\{\phi\}_r^T$: $$ \{\phi\}_s^T[K]\{\phi\}_r = \omega_r^2\{\phi\}_s^T[M]\{\phi\}_r, \qquad \{\phi\}_r^T[K]\{\phi\}_s = \omega_s^2\{\phi\}_r^T[M]\{\phi\}_s $$ Since $[K]$ and $[M]$ are symmetric, the left-hand sides are equal (transpose of a scalar) and so are the mass products. Subtracting: $$ (\omega_r^2 - \omega_s^2)\{\phi\}_r^T[M]\{\phi\}_s = 0 \;\Rightarrow\; \{\phi\}_r^T[M]\{\phi\}_s = 0 \quad (r \ne s) $$ and hence also $\{\phi\}_r^T[K]\{\phi\}_s = 0$ for $r \ne s$. For $r = s$ the products are the **modal mass** $M_r = \{\phi\}_r^T[M]\{\phi\}_r$ and **modal stiffness** $K_r = \{\phi\}_r^T[K]\{\phi\}_r = \omega_r^2M_r$. Normalising so that $M_r = 1$ (mass-normalised) gives $[\Phi]^T[M][\Phi] = [I]$ and $[\Phi]^T[K][\Phi] = \text{diag}(\omega_r^2)$. #### Modal analysis for forced response For $[M]\{\ddot{x}\} + [K]\{x\} = \{F(t)\}$: 1. Find the natural frequencies $\omega_r$ and the mode-shape matrix $[\Phi]$ (columns normalised to unit modal mass). 2. Use the transformation $\{x\} = [\Phi]\{q\}$, where $\{q\}$ are the principal coordinates. 3. Substitute and pre-multiply by $[\Phi]^T$. By orthogonality the equations uncouple: $$ \ddot{q}_r + \omega_r^2q_r = Q_r(t), \qquad Q_r(t) = \{\phi\}_r^T\{F(t)\} $$ 4. Solve each single-degree-of-freedom equation, for example with the Duhamel (convolution) integral: $$ q_r(t) = q_r(0)\cos\omega_rt + \frac{\dot{q}_r(0)}{\omega_r}\sin\omega_rt + \frac{1}{\omega_r}\int_0^tQ_r(\tau)\sin\omega_r(t-\tau)\,d\tau $$ 5. Transform back: $\{x(t)\} = [\Phi]\{q(t)\} = \sum_r\{\phi\}_rq_r(t)$. For harmonic force $Q_r\sin\omega t$, the steady-state result is $q_r = \dfrac{Q_r}{\omega_r^2 - \omega^2}\sin\omega t$. Often only the first few modes are needed, which reduces the size of the problem greatly. With proportional damping ($[C] = \alpha[M] + \beta[K]$) the same method works with a damping ratio $\zeta_r$ for each mode. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the proof, pre-multiplying each eigen-equation by the other mode's transpose and subtracting gives: A. (w_r^2 - w_s^2) {phi}r^T [M] {phi}s = 0 B. (w_r + w_s) {phi}r^T {phi}s = 0 C. {phi}r^T [M] {phi}r = 0 D. w_r^2 = w_s^2 Answer: A. Symmetry of [K] and [M] cancels the stiffness terms; with w_r not equal to w_s the mass product must vanish. (Key point: Orthogonality proof step) 2. The conclusion {phi}r^T[M]{phi}s = 0 needs: A. symmetric [M] and [K] and w_r not equal to w_s B. equal natural frequencies C. diagonal [K] D. zero damping only in one mode Answer: A. Symmetry makes the stiffness terms equal on subtraction, and w_r not equal to w_s lets us divide out the factor. (Key point: Role of symmetry and distinct frequencies) 3. For r = s, the modal stiffness K_r equals: A. M_r / w_r^2 B. w_r M_r C. w_r^2 M_r D. M_r^2 Answer: C. From [K]phi = w^2 [M]phi, K_r = phi^T K phi = w_r^2 M_r. (Key point: Modal mass and stiffness) 4. Using x = [Phi]q and pre-multiplying by [Phi]^T gives for each mode: A. a coupled set of n equations B. q_r'' + w_r^2 q_r = Q_r(t), with Q_r = phi_r^T F(t) C. q_r'' + q_r = F(t) D. an algebraic equation with no time dependence Answer: B. Orthogonality diagonalises M and K so each principal coordinate obeys an independent SDOF equation. (Key point: Uncoupling the equations) 5. After solving each q_r(t) (e.g. with Duhamel's integral), the physical response is: A. x = q_1 only B. x = sum of phi_r q_r(t) over modes C. x = product of q_r D. x = [Phi]^-1 F Answer: B. {x(t)} = [Phi]{q(t)} = sum {phi}r q_r(t); often only the first few modes are needed. 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